Why differential of e^x is special?

[fission]

I learned that differential of e^x is same but what's so special about it? What makes is so special as it seems like a normal function to me other than the fact that e= sum of series of reciprocal of factorial numbers. What i want to ask is if e^x differential is e^x then do this rule apply to b^x too? Where b is a positive base. Why or why not?

 

[member 587159]

I learned that differential of e^x is same but what's so special about it? What makes is so special as it seems like a normal function to me other than the fact that e= sum of series of reciprocal of factorial numbers. What i want to ask is if e^x differential is e^x then do this rule apply to b^x too? Where b is a positive base. Why or why not?

No $\frac{db^x}{dx} = \log_e (b) b^x$.
So we get the desired formula when $b = e$

When you would look at differentiation as an operation in a function space , $e^x$ would be a neutral element for this operation.

 

[Mark44]

I learned that differential of e^x is same but what's so special about it? What makes is so special as it seems like a normal function to me other than the fact that e= sum of series of reciprocal of factorial numbers. What i want to ask is if e^x differential is e^x then do this rule apply to b^x too? Where b is a positive base. Why or why not?

Actually, what you're asking about is the derivative of e^x, not the differential of e^x.

 

[fresh_42]

I learned that differential of e^x is same but what's so special about it? What makes is so special as it seems like a normal function to me other than the fact that e= sum of series of reciprocal of factorial numbers. What i want to ask is if e^x differential is e^x then do this rule apply to b^x too? Where b is a positive base. Why or why not?

One can ask which functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ satisfy $\frac{d}{dx}f(x) = f(x)$. The solution is $x \mapsto c \cdot e^x$.
So $x \mapsto e^x$ is the solution that additionally satisfies the condition $f(0) = 1$.
This is one way to define the exponential function.

Others are a limit or a version of the series you mentioned.

As you also mentioned $g(x) = b^x$, there is one thing they have in common: both satisfy $g(x+y) = g(x) + g(y)$.
But as @Math_QED has said, $\frac{d}{dx} g(x) = \log(b) \cdot g(x) \neq g(x)$ if the basis isn't $e$.